921 Minimum Add to Make Parentheses Valid

1 题目

Given a string S of '(' and ')' parentheses, we add the minimum number of parentheses ( '(' or ')', and in any positions ) so that the resulting parentheses string is valid.

Formally, a parentheses string is valid if and only if:

• It is the empty string, or
• It can be written as AB (A concatenated with B), where A and B are valid strings, or
• It can be written as (A), where A is a valid string.

Given a parentheses string, return the minimum number of parentheses we must add to make the resulting string valid.

Example 1:

Input: "())"
Output: 1

 Example 2:

Input: "((("
Output: 3

Example 3:

Input: "()"
Output: 0

Example 4:

Input: "()))(("
Output: 4

Note:

1. S.length <= 1000
2. S only consists of '(' and ')' characters.

2 尝试解

2.1 分析

给定一串由"(",")"组成的字符串，但是括号的组合不一定有效。问至少要添加多少个"("或")"才能使得该字符串是有效的括号组合。

用left记录左括号的个数，result记录需要添加的括号个数。

如果遇到右括号，且左括号有剩余，则右括号与最近的一个左括号配对,left--；如果左括号没有剩余，则需要添加一个左括号。

如果遇到左括号，则left++。最后左括号有剩余，则需要给他们配上相同数目的右括号，result+=left。

2.2 代码

class Solution {
public:
    int minAddToMakeValid(string S) {
        int result = 0;
        int left = 0;
        for(auto letter : S){
            if(letter == '(')
                left++;
            else{
                if(left > 0) left--;
                else result += 1;
            }
        }
        result += left;
        return result;
    }
};

3 标准解

class Solution {
public:
    int minAddToMakeValid(string S) {
        int left = 0, right = 0;
        for (char c : S)
            if (c == '(')
                right++;
            else if (right > 0)
                right--;
            else
                left++;
        return left + right;
    }
};